Proving that $mathbf{(H-frac{1}{n}J_n)}$ is indempotent











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I am trying to show that the matrix $mathbf{(H-frac{1}{n}J_n)}$ is idempotent where $mathbf{H}$ is the Hat-matrix (Projection matrix) of linear regression and $J_n$ is the $ntimes n$ matrix with $1$ in all its inputs. Taking :



$$mathbf{(H-frac{1}{n}J_n)(H-frac{1}{n}J_n)= HH - Hfrac{1}{n}J_n - frac{1}{n}J_nH + frac{1}{n}J_nfrac{1}{n}J_n}$$



Now, we know that $mathbf{H}$ and $mathbf{frac{1}{n}J_n}$ are idempontent, thus :



$$mathbf{(H-frac{1}{n}J_n)(H-frac{1}{n}J_n)=H-Hfrac{1}{n}J_n - frac{1}{n}J_nH +frac{1}{n}J_n}$$



How would I continue now in order to show that the given matrix is idempontent ?






linear-algebra matrices linear-regression idempotents







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  • The "hat matrix" should be redefined, It is not a usual name...
    – Jean Marie
    Nov 22 at 19:08










  • @JeanMarie It is the projection matrix. en.wikipedia.org/wiki/Projection_matrix
    – Rebellos
    Nov 22 at 19:12








  • 1




    @JeanMarie It's $X(X^TX)^{-1}X^T$, where $X$ is the matrix of regressors. It's called the hat matrix because when you multiply $y$ on the left by it, you get $hat y$, that is, it "puts a hat on $y$", figuratively.
    – Jean-Claude Arbaut
    Nov 22 at 19:15








  • 1




    @Jean-Claude Arbaut Thank you for this precise explanation. Thank you as well to the OP.
    – Jean Marie
    Nov 22 at 19:19










  • Looks to me $H$ has to be the identity matrix for $H-frac{1}{n}J_n$ to be idempotent.
    – StubbornAtom
    Nov 22 at 19:26















up vote
1
down vote

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I am trying to show that the matrix $mathbf{(H-frac{1}{n}J_n)}$ is idempotent where $mathbf{H}$ is the Hat-matrix (Projection matrix) of linear regression and $J_n$ is the $ntimes n$ matrix with $1$ in all its inputs. Taking :



$$mathbf{(H-frac{1}{n}J_n)(H-frac{1}{n}J_n)= HH - Hfrac{1}{n}J_n - frac{1}{n}J_nH + frac{1}{n}J_nfrac{1}{n}J_n}$$



Now, we know that $mathbf{H}$ and $mathbf{frac{1}{n}J_n}$ are idempontent, thus :



$$mathbf{(H-frac{1}{n}J_n)(H-frac{1}{n}J_n)=H-Hfrac{1}{n}J_n - frac{1}{n}J_nH +frac{1}{n}J_n}$$



How would I continue now in order to show that the given matrix is idempontent ?






linear-algebra matrices linear-regression idempotents







share|cite|improve this question
























  • The "hat matrix" should be redefined, It is not a usual name...
    – Jean Marie
    Nov 22 at 19:08










  • @JeanMarie It is the projection matrix. en.wikipedia.org/wiki/Projection_matrix
    – Rebellos
    Nov 22 at 19:12








  • 1




    @JeanMarie It's $X(X^TX)^{-1}X^T$, where $X$ is the matrix of regressors. It's called the hat matrix because when you multiply $y$ on the left by it, you get $hat y$, that is, it "puts a hat on $y$", figuratively.
    – Jean-Claude Arbaut
    Nov 22 at 19:15








  • 1




    @Jean-Claude Arbaut Thank you for this precise explanation. Thank you as well to the OP.
    – Jean Marie
    Nov 22 at 19:19










  • Looks to me $H$ has to be the identity matrix for $H-frac{1}{n}J_n$ to be idempotent.
    – StubbornAtom
    Nov 22 at 19:26













up vote
1
down vote

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1









up vote
1
down vote

favorite
1






1





I am trying to show that the matrix $mathbf{(H-frac{1}{n}J_n)}$ is idempotent where $mathbf{H}$ is the Hat-matrix (Projection matrix) of linear regression and $J_n$ is the $ntimes n$ matrix with $1$ in all its inputs. Taking :



$$mathbf{(H-frac{1}{n}J_n)(H-frac{1}{n}J_n)= HH - Hfrac{1}{n}J_n - frac{1}{n}J_nH + frac{1}{n}J_nfrac{1}{n}J_n}$$



Now, we know that $mathbf{H}$ and $mathbf{frac{1}{n}J_n}$ are idempontent, thus :



$$mathbf{(H-frac{1}{n}J_n)(H-frac{1}{n}J_n)=H-Hfrac{1}{n}J_n - frac{1}{n}J_nH +frac{1}{n}J_n}$$



How would I continue now in order to show that the given matrix is idempontent ?






linear-algebra matrices linear-regression idempotents







share|cite|improve this question















I am trying to show that the matrix $mathbf{(H-frac{1}{n}J_n)}$ is idempotent where $mathbf{H}$ is the Hat-matrix (Projection matrix) of linear regression and $J_n$ is the $ntimes n$ matrix with $1$ in all its inputs. Taking :



$$mathbf{(H-frac{1}{n}J_n)(H-frac{1}{n}J_n)= HH - Hfrac{1}{n}J_n - frac{1}{n}J_nH + frac{1}{n}J_nfrac{1}{n}J_n}$$



Now, we know that $mathbf{H}$ and $mathbf{frac{1}{n}J_n}$ are idempontent, thus :



$$mathbf{(H-frac{1}{n}J_n)(H-frac{1}{n}J_n)=H-Hfrac{1}{n}J_n - frac{1}{n}J_nH +frac{1}{n}J_n}$$



How would I continue now in order to show that the given matrix is idempontent ?






linear-algebra matrices linear-regression idempotents




linear-algebra matrices linear-regression idempotents






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 22 at 19:13

























asked Nov 22 at 19:05









Rebellos

13.9k31243




13.9k31243












  • The "hat matrix" should be redefined, It is not a usual name...
    – Jean Marie
    Nov 22 at 19:08










  • @JeanMarie It is the projection matrix. en.wikipedia.org/wiki/Projection_matrix
    – Rebellos
    Nov 22 at 19:12








  • 1




    @JeanMarie It's $X(X^TX)^{-1}X^T$, where $X$ is the matrix of regressors. It's called the hat matrix because when you multiply $y$ on the left by it, you get $hat y$, that is, it "puts a hat on $y$", figuratively.
    – Jean-Claude Arbaut
    Nov 22 at 19:15








  • 1




    @Jean-Claude Arbaut Thank you for this precise explanation. Thank you as well to the OP.
    – Jean Marie
    Nov 22 at 19:19










  • Looks to me $H$ has to be the identity matrix for $H-frac{1}{n}J_n$ to be idempotent.
    – StubbornAtom
    Nov 22 at 19:26


















  • The "hat matrix" should be redefined, It is not a usual name...
    – Jean Marie
    Nov 22 at 19:08










  • @JeanMarie It is the projection matrix. en.wikipedia.org/wiki/Projection_matrix
    – Rebellos
    Nov 22 at 19:12








  • 1




    @JeanMarie It's $X(X^TX)^{-1}X^T$, where $X$ is the matrix of regressors. It's called the hat matrix because when you multiply $y$ on the left by it, you get $hat y$, that is, it "puts a hat on $y$", figuratively.
    – Jean-Claude Arbaut
    Nov 22 at 19:15








  • 1




    @Jean-Claude Arbaut Thank you for this precise explanation. Thank you as well to the OP.
    – Jean Marie
    Nov 22 at 19:19










  • Looks to me $H$ has to be the identity matrix for $H-frac{1}{n}J_n$ to be idempotent.
    – StubbornAtom
    Nov 22 at 19:26
















The "hat matrix" should be redefined, It is not a usual name...
– Jean Marie
Nov 22 at 19:08




The "hat matrix" should be redefined, It is not a usual name...
– Jean Marie
Nov 22 at 19:08












@JeanMarie It is the projection matrix. en.wikipedia.org/wiki/Projection_matrix
– Rebellos
Nov 22 at 19:12






@JeanMarie It is the projection matrix. en.wikipedia.org/wiki/Projection_matrix
– Rebellos
Nov 22 at 19:12






1




1




@JeanMarie It's $X(X^TX)^{-1}X^T$, where $X$ is the matrix of regressors. It's called the hat matrix because when you multiply $y$ on the left by it, you get $hat y$, that is, it "puts a hat on $y$", figuratively.
– Jean-Claude Arbaut
Nov 22 at 19:15






@JeanMarie It's $X(X^TX)^{-1}X^T$, where $X$ is the matrix of regressors. It's called the hat matrix because when you multiply $y$ on the left by it, you get $hat y$, that is, it "puts a hat on $y$", figuratively.
– Jean-Claude Arbaut
Nov 22 at 19:15






1




1




@Jean-Claude Arbaut Thank you for this precise explanation. Thank you as well to the OP.
– Jean Marie
Nov 22 at 19:19




@Jean-Claude Arbaut Thank you for this precise explanation. Thank you as well to the OP.
– Jean Marie
Nov 22 at 19:19












Looks to me $H$ has to be the identity matrix for $H-frac{1}{n}J_n$ to be idempotent.
– StubbornAtom
Nov 22 at 19:26




Looks to me $H$ has to be the identity matrix for $H-frac{1}{n}J_n$ to be idempotent.
– StubbornAtom
Nov 22 at 19:26















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